Question

A circular plate of uniform thickness has a diameter of $28\text{cm}$. A circular portion of diameter $21\text{cm}$ is removed from the plate as shown. $O$ is the centre of mass of complete plate. The position of centre of mass of remaining portion will shift towards left from $O$ by

• A. $5\text{cm}$
• B. $9\text{cm}$
• C. $4.5\text{cm}$
• D. $5.5\text{cm}$

Answer 1

The correct option is C $4.5\text{cm}$



$C_1$ is the cenre of mass of cut portion and $C_2$ that of remaining portion. We have to find $x_2.$
$x_1=14-10.5=3.5\text{cm}$
Mass will be proportional to area. So mass of the whole disc is
$M=k\pi (14{)}^2$
Mass of cut portion, $m_1=k\pi (10.5{)}^2$
Mass of the remaining portion $m_2=M-m_1=k\pi ({14}^2-{10.5}^2)=k\pi \left(24.5\right)\times \left(3.5\right)$
Now, $m_1{x}_1=m_2{x}_2$
$\Rightarrow x_2=\frac{m_1{x}_1}{m_2}=\frac{k\pi (10.5{)}^2\times 3.5}{k\pi \left(24.5\right)\times 3.5}=4.5\text{cm}$